It means that, even when u know the eigenvalues of all the compatible observables after measurement, u cannot use that to determine the measured state of the system becuz it cud b any linear combination. However in reality, particles r identical, so there is not much choice in the ket. I think this is what the book means by "nature avoiding this difficulty"
It means that only that those kets r measured that r symmetric or antisymmetric under exchange of the eigenvalues
Becuz this means particles r identical
@SillyGoose I dont think this argument always works becuz the total hamiltonian also has an extra interaction term, apart from being the sum of free Hamiltonians
But this argument does work 4 angular momentum. Becuz adding the individual operators means the new operators also satisfy the angular momentum commutation relations. So they make sense as generators
More generally, the observables of the composite system can b arrived at using Noether's theorem applied to the composite system's translation and rotation symmetries
But this too doesnt always work becuz systems without symmetries also have observables
Generally, i think u can first define space-time translations and rotation transformations of a system first (regardless of whether they're symmetries or not). And then the generators of those r defines to be observables
@SillyGoose But you are right that if two systems each carry a representation of the Poincaire group, then the composite system can trivially be given a representation of the Poincaire group by merely adding the generators from the two systems. So this is y addition is fine mathematically. But it may not work physically, becuz the physical Hamiltonian need not b a sum of the individual Hamiltonians
This addition method would b physically correct for non-interacting systems
@SillyGoose There's a general principle here: The subsystems are composed via the tensor product, $H_1\otimes H_2$. Unitary operators acting on these states are likewise composed with the tensor product, $U_1\otimes U_2$. But the self-adjoint operators $T_1,T_2$ as generators of unitaries $U_i = \mathrm{e}^{\mathrm{i}\epsilon_i T_i}$ - because the exponential turns addition into multiplication - then compose via addition, $T_1\oplus T_2 = T_1 \otimes \mathbf{1} + \mathbf{1}\otimes T_2$
@RyderRude i see the hamiltonian counterexample does make sense… about the poincare group business, adding the generators just amounts to applying the same transformation on both subsystems, right? Is this what is meantby it is trivial
I see—i was trying to make a connection between the direct sum structure of Herman operators and the tensor product structure of the unitary operator and what that means in practice. But now i am confused why the hamiltonian doesn’t fall under this general behavior? Or should we really consider the hamiltonian as a context dependent thing. (Also i mean to only talk about two-part identical systems)
abstractly, all that's happening here is that products of Lie groups $G\times H$ act on tensor product representations $V_G\otimes V_H$ and their Lie algebra is $\mathfrak{g}\oplus\mathfrak{h}$
@SillyGoose I mean, a free Hamiltonian acts exactly like that
if your two subsystems don't interact, then the total Hamiltonian is just the sum of the subsystem Hamiltonians
but usually there is some interaction term in there otherwise you'd not consider the two systems as "subsystems" of a larger system, i.e. the reason you're thinking about this in the first place is that your Hamiltonian doesn't factorize
well i guess now i am thinking that the hamiltonian really does seem singled out from all other observables in a nontrivial way, namely, because it can have interaction terms
i mean couldn't one say we should say the translation and rotation unitary operators togetehr are special because they give rise to dynamics wrt to space
I'm not sure what you're after here - the thing the generates time evolution is "special" because more or less the whole reason we do physics is to predict time evolutions of systems
when I throw a ball I want to know where it will be in a few seconds, not where it would be if someone made a spatial translation on it :P
isn't it slices of spacetime we are studying? the thing is that with a single time variable we can study the evolution of an entire "frame", but a single space variable doesn't uniquely identify a frame, does it?
The nice thing about a single time dimension is that the Cauchy problem is - under relatively mild assumptions - well-posed: The data on a single spatial slice + knowledge of the dynamics allows you to predict everything
in more than one time dimension the data on a space-like slice no longer suffices
i.e. if we start with (---+) then going to five dimensions where it is (---++) we've added a time dimension and if its (----+) we've added a space dimension
there is F-theory which formally works in 12d, two of which are time dimensions but that's a very special case and the physics you actually do is after compactifying one of the time dimensions and one of the space dimensions on a torus
can i think of the permutation (symmetry) unitary operator $P_{12}$ as 1) a map between operator algebras of "identical HS"; $P_{12}: \mathcal{A}(\mathcal{H}_1) \rightarrow \mathcal{A} (\mathcal{H}_2)$ as well as 2) a permutation $P_{12}: \mathcal{A}(\mathcal{H}_1) \rightarrow \mathcal{A} (\mathcal{H}_1)$ (or $\mathcal{H}_2$)
err i mean is perspective 1) just essentially what is happening
I don't know how you think either of these maps work
the notion of permutation only makes sense when $H_1 = H_2 = H$. Then permutation is $A(H)\otimes A(H) \mapsto A(H)\otimes A(H), X\otimes Y \mapsto Y\otimes X$
fqq already said it: The states transform in a unitary representation of the Lorentz group, while we usually talk about tensor powers of the fundamental finite-dimensional representation of the Lorentz group
You will have $U(\Lambda)\lvert p^\mu \rangle = \lvert \Lambda^\mu_\nu p^\nu\rangle$, but that doesn't fit with what we usually mean by the state "carrying a Lorentz index"
because that would have the $\Lambda^\mu_\nu$ acting on the state itself
the finite-dimensional representation is on the target space of the fields
A vector field $A_\mu$ is a map into the finite-dimensional space $\mathbb{R}^4$, on which the fundamental rep of $\mathrm{SO}(1,3)$ acts
and also, you should not contrast this with classical mechanics, but with non-relativistic QM: There, the symmetry group is $\mathrm{SO}(3)$, and all the unitary reps of $\mathrm{SO}(3)$ are finite-dimensional - the spin spaces of dimension $2s+1$
Just so I understand, RQM is the same as classical field theory in this respect (i.e. the wavefunction is essentially a field that obeys the Schrodinger equation)
well, then it will be additionally valued in the finite-dimensional spin-1 rep of so(1,3), and transform as $\psi^\mu(x) \mapsto \Lambda^\mu_\nu \psi(\Lambda x)$
the point is that when we don't use some concrete representations as functions of position - and we have discussed ad nauseam why position is "bad" - then $\psi(x)\mapsto \psi(\Lambda x)$ becomes $\lvert \psi\rangle \mapsto U(\Lambda)\lvert \psi\rangle$ in an abstract notation, and that $U(\Lambda)$ is infinite-dimensional
contrast this, again, with non-rel. QM where you also have rotations operating as $\psi(x) \mapsto \psi(Rx)$ on an infinite-dimensional function space, but where the space of states decomposes into finite-dimensional irreducible representations of SO(3) - the spherical harmonics
Why would there be linear representations on a classical state space? Classically, the space of states has neither a vector space structure nor an inner product, so the notion of linear representation, let alone unitary representation, does not make sense
you can prove that there are no finite-dimensional unitary representations of the Lorentz group
essentially because it is non-compact but all finite-dimensional unitary groups are compact, so it's impossible to construct a Lie group homomorphism from the Lorentz group into any finite-dimensional unitary group
@SillyGoose u can also think of it like this : rotations and translations have an expected behavior. Like rotations are periodic and translations r commutative and stuff. Their group structure is fixed. If our definition of rotation and translation did not have this behavior, we wudnt call them rotations and translations in the first place
But there is lot of freedom in defining the time evolution function. There is no prior expectation in how it should behave
U can even have time - dependent Hamiltonians. It would still work as a notion of time evolution
Such systems would not have time translation symmetry. But even those systems would have rotations and translations behaving in the expected way
But i dont think this reasoning is very good.. Becuz requiring the Poincaire group does put a restriction on the behavior of Hamiltonian
Yeah it is costly. But since again personally I bought it to read books, I figured it would quickly cover the cost of all the physical books I would have ordered in some reasonable interval of time :)
It's too slow and it's only useful to write (which my tab s6 can do better with respect to any feature except the feeling and the paperlike screen)
I can't use it to read pdf's and write things at the same time (with normal pace)
@Amit It isn't bad to have per se, just not suitable for what I need
But I think we're on the right path: the feeling is very good and it's comfortable to write on it even on a damn bus. I hope they'll make e-ink tablets with better software and solve the slow-screen "problem" due to the current e-ink technology
Well I would have bought a lot more physics/math kindle books for it if not for the annoying reality I discovered, that the kindle version of most of these types of books is worse than just a plain PDF! You know why? Because the lack of compatibility of the mathematical expressions with kindle turns out to make them irreparably tiny lol. I say irreparably because as opposed to PDF you have less freedom to mess with the kindle format, probably due to DRM issues.
I didn't experience slowness tbh... but then as far as speed goes maybe I knew what to expect. Where do you find it too slow?
My problem with PDFs is not the slowness, it's just that again PDFs are for A4 size... so every PDF page needs to be read as two "pieces" on this device, which is annoying sometimes when you need to back reference something
there is nothing inherently A4 to PDFs, it's just that many lecture notes and articles are written in A4 format because they're meant to be printed in A4
there are also plenty of PDFs around for the slightly different US Letter format, for example
@Amit no, not really. I spotchecked two Springer textbook PDFs I have and they're both 155x235 mm, which is "large textbook format" and the size they would be actually printed at but smaller than DIN A4 (210x297 mm)
@ACuriousMind Interesting, but did you purchase those files? I think that maybe the various commercial eBook type files may indeed have a size more suitable for an eReader?
Yes, I didn't mean that eBook is a specific format, just eBook as a commercial term. I am just speculating that perhaps the purchased ones are more likely to be better suited for a device than for printing on A4, hence the smaller page size
@Amit I'm not saying you can't read a PDF on an eBook reader, I'm saying the experience of the reader is usually optimized to the format in which you'd purchase a book from its standard marketplace, which is not PDF in any case I know (and usually .epub)
in particular PDFs have defined pages, while the readers are really optimized for you to adjust your preferred text size and then display "pages" of whatever size comfortably fits on screen in that font size
@ACuriousMind Well, we can do a simple experiment! If you name of one of the books you purchased, we can check to see if the copies we could just "find" of them online, are A4 or not
@ACuriousMind Yes, that is an important flexibility. But even just having a smaller page size already makes the book a lot more readable on many readers. Also note that a lot of devices can do a certain amount of PDF rearranging dynamically (I know mine can) -- doesn't always work perfectly but I found it useful on occasion
Do the universities have deals with the academic publishers? So if a certain researcher wants to upload his papers to arxiv he can't necessarily do it?
that's why it's a "pre-print server" - technically the versions there are supposed to be preliminary versions, while the published paper is supposed to be more polished
in practice that's not what usually happens but everyone kinda just ignores it
also note that this is very much field dependent - while math and physics have almost universally established a culture of uploading pre-prints, this isn't universal across all academic fields and not even all STEM fields
Well now I finally understand why it has to be under the pretense of "preprint" - the first place I saw this term is IACR, same idea as arxiv but for Cryptography papers
I'm also bored with 99% of the marvel stuff. In fact guardians of the galaxy is significantly different somehow for my taste, maybe because the actors are funnier
I don't want more book adaptations, either. Make something original! I really dislike this trend of franchising and rebooting, which mostly is just done because they can reap an existing fan base
Actually the superhero stuff is more fantasy than sci-fi for sure, but where it fails I think is that it takes a very very special mixture of actors and writing to translate a good comic into a good movie.... and when they started they did it very well, but now they are just feeding off past successes with mediocre movies
I think adaptations can be very good, it's a matter of "implementation"... I mean yeah it is kind of cheap in the sense of reaping an existing fan base, but if you want to create a franchise you want people to come back for more than one movie... and that means probably you have to appeal not only to the current fan base but beyond that
@RyderRude Okay, I'm not only bored of the franchises, I'm also bored of everyone discussing them. If they just existed and I could ignore them I probably would be indifferent
@RyderRude well, but generally you can just avoid specific things if you don't like them. These large franchises have invaded pop culture mainstream thoroughly and since they keep making new ones the discussion never stops
Maybe adaptation stuff should be a bit more like James Bond -- start as faithful to the books, but then break free from that and write stuff which is better tailored to the screen
What is the basis for the representation space of $\hat{U}$? is it $|p_1^\mu \rangle,\,|p_2^\mu \rangle,\dots$?
@ACuriousMind In non-rel is it just $e^{i k x}$ for all different $k$? Are we saying infinite dimensional because we have a basis consisting of infinite elements?
@WaveInPlace always edit (unless the edit changes the question so much it would invalidate existing answers, in which case you probably actually want to ask a new, more focused question)
